91Ó°ÊÓ

1. The molar mass of iodine is,
   $M=127\text{g}/\text{mol}=127×{10}^{-3}\text{kg}/\text{mol}$
2. The frequency of sound is$f=1000\text{Hz}$.
3. The temperature is $T=400\text{K}$.
4. The nodes of the standing wave are separated by 9.57 cm.

By using the equation for the speed of sound in gas from the problems 19-91, andthe velocity of sound wave from Eq. (16-13), find$\mathrm{γ}$for the gas.

${\mathbf{v}}_{\mathbf{s}}\mathbf{=}\sqrt{\frac{\mathbf{γ¸é°Õ}}{\mathbf{M}}}$

${\mathbf{v}}_{\mathbf{s}}\mathbf{=}\mathbf{λ´Ú}$

where, Ris the gas constant,Tis the temperature, vis velocity, is wavelength, f is frequency andMis the molar mass.

From problems 19-91, the speed of the sound in the gas is,

$v_s=\sqrt{\frac{\mathrm{γ}RT}{M}}$

Where$\mathrm{γ}={\text{C}}_p/{\text{C}}_V$ ,T is the temperature ,R is gas constant, and Mis the molar mass.

Squaring both sides,

$v_s^2=\frac{\mathrm{γ}RT}{M}$

Therefore,

$\mathrm{γ}=\frac{M{v}_s^2}{RT}………………………………………….\left(1\right)$

From Eq. (16-13) the velocity of sound wave is,

$v_s=\mathrm{λ}f……………………………………\left(16-13\right)$

Since the nodes of the standing wave are separated by half a wavelength, the wavelength is given by,

$$\begin{gathered} \mathrm{λ}=2×\left(9.57\text{cm}\right) \\ =19.14\text{cm} \\ =0.1914\text{m} \end{gathered}$$

Putting in Eq. (16-13),

$$\begin{gathered} v_s=0.1914\text{m}×1000\text{Hz} \\ =191.4\text{m}/\text{s} \end{gathered}$$

Putting this in Eq. (1),

$$\begin{gathered} \mathrm{γ}=\frac{127×{10}^{-3}\text{g}/\text{mol}×{\left(191.4\text{m}/\text{s}\right)}^2}{8.31\text{J}/\text{mol}.\text{k}×400\text{K}} \\ =1.40 \end{gathered}$$

Hence, $\mathrm{γ}$for the gas is $\mathrm{γ}=1.40$.

By using the equation for the speed of sound in gas from the problems 19-91, andthe equation for the velocity of sound wave, this problem can be solved.